$\sqrt{3} \sec x+\csc x+2(\tan x-\cot x)=0$

$\Rightarrow \sqrt{3} \sec x+\csc x=2(\cot x-\tan x)$

$\Rightarrow \frac{(\sqrt{3} \sec x+\csc x)}{2}=\cot x-\tan x$

$\Rightarrow \frac{\sqrt{3}}{2} \sec x +\frac{1}{2} \csc x =\cot x -\tan x$ by dividing both sides by $2$

We know that $\sec x=\frac{1}{\cos x}, \csc x=\frac{1}{\sin x}, \tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$ $\Rightarrow \frac{\sqrt{3} 1}{2 \cos x}+\frac{11}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$

$\Rightarrow \frac{\sqrt{3} \sin x}{2 \sin x \cos x}+\frac{1 \cos x}{2 \sin x \cos x}=\frac{\cos ^2 x}{\sin x \cos x}-\frac{\sin ^2 x}{\cos x \sin x}$

$\Rightarrow \frac{\sqrt{3}}{2} \sin x +\frac{1}{2} \cos x =\cos ^2 x -\sin ^2 x$

$\Rightarrow \sin \frac{\pi}{3} \sin x+\cos \frac{\pi}{3} \cos x=\cos 2 x$ where $\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}$ and $\frac{1}{2}=\cos \frac{\pi}{3}$

$\Rightarrow \cos \left(\frac{\pi}{3}-x\right)=\cos 2 x$

$\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$ since if $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha$

$\Rightarrow 2 x =2 n \pi+ x -\frac{\pi}{3}$ or $2 x =2 n \pi- x +\frac{\pi}{3}$

$\Rightarrow x=2 n \pi-\frac{\pi}{3}$ or $3 x=2 n \pi+\frac{\pi}{3}$ or $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$

For $n =0, x =\frac{-\pi}{3}, \frac{\pi}{9}$

For $n =-1, x =\frac{-7 \pi}{3}$ which does not lie between $(-\pi, \pi)$

and $x=\frac{-2 \pi}{3}+\frac{\pi}{9}=\frac{-5 \pi}{9} \in(-\pi, \pi)$

For $n =2, x =\frac{5 \pi}{3}$ does not lie between $(-\pi, \pi)$

and $x=\frac{2 \pi}{3}+\frac{\pi}{9}=\frac{7 \pi}{9} \in(-\pi, \pi)$

$\therefore x =\frac{-\pi}{3}, \frac{\pi}{9}, \frac{-5 \pi}{9}, \frac{7 \pi}{9}$

Sum of all the solutions $=\frac{-\pi}{3}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=\frac{-3 \pi}{9}+\frac{\pi}{9}-\frac{5 \pi}{9}+\frac{7 \pi}{9}=0$